Here is a tedious & unsatisfactory answer. I suspect there is a simpler answer, but it evades me. It is unsatisfactory because it relies on
the Faà di Bruno formula (in particular, see the Michael Hardy
reference)
for derivatives of compositions. This is unsatisfactory purely because it
is a little opaque, and not part of everybody's bag of tricks. I also
use multi index notation, mainly to preserve my notational sanity.
The proof follows the following lines:
- Reduce the problem to showing that $\rho(x)= r(\|x\|)$ is $C^k$,
where $r^{(i)}(0) = 0$, for $i=0,...,k$.
- Find a suitable bound for the derivatives of $x \mapsto \|x\|$, for $x \neq 0$.
- Show the relevant derivatives of $\rho$, for $x \neq 0$, converge to $0$ as $x \to 0$.
- Show the relevant derivatives of $\rho$ are differentiable
at $x=0$ and that the derivatives are zero.
First a few notes:
Note that if a $C^1$ function $g$ is even, then $g'$ is odd and hence $g'(0) = 0$. If a $C^1$ function $g$ is odd, then $g'$ is even. Hence if $g$ is an even $C^k$ function, then the odd derivatives satisfy $g^{2i+1}(0) = 0$.
Let $n(x) = \|x\|$, we note that $n$ is smooth for $x \neq 0$, and $n^2$ is smooth everywhere.
Let $\phi = f \circ n$. If $f$ is $C^k$, then $\phi $ is $C^k$ for $x \neq 0$. Hence the only issue here is to show that $\phi$ is $C^k$ at zero (meaning the relevant derivatives exist at zero and are continuous there).
Now the reduction:
Since $f$ is $C^k$ and even, Taylor's theorem shows that there is some that $C^k$ function $r$ such that
$f(t) = \sum_{i=0, \text{even}}^k {f^{(i)}(0) \over i! } t^i + r(t)$, and $r$
satisfies $r^{(i)}(0) = 0$ and $\lim_{t \to 0}{r^{(i)}(t) \over t^{k-i}} = 0$ for $i=0,...,k$. Since the Taylor polynomial only involves even terms, and $n^2$ is smooth, we see that showing $\phi$ is $C^k$ is equivalent to showing
that $\rho=r \circ n$ is $C^k$. So, we need to show that
$\rho$ is $C^k$. In light of the above comments, we need only deal with
the point $x=0$.
We will need a technical bound on $r$: Note that for any $\epsilon>0$ there is some $\delta \in (0,1]$ such that if $|t|< \delta$, then
$|r^{(i)}(t)| \le \epsilon | t^{k-i} | $, for all $i=0,...,k$.
Now the tedium:
Suppose $x \neq 0$. We have
${\partial n(x) \over \partial x_i} = {x_i \over \|x\|}$ for all $i$. Now let
$\alpha$ be a multi index, then I claim that
${\partial^\alpha n(x) \over \partial x^\alpha}$ has the form
${\partial^\alpha n(x) \over \partial x^\alpha} = {p_\alpha(x) \over \|x\|^{2 |\alpha|-1}}$ for some polynomial $p_\alpha$, that satisfies $\partial p_\alpha = |\alpha|$ (that is the degree is $\alpha$) and $p_\alpha(x) = 0$. Note that
${\partial^\alpha n(x) \over \partial x^\alpha} = {\partial^\beta n(x) \over \partial x^\beta}$ if $\beta$ is a permutation of $\alpha$ since $n$ is smooth (in particular, the order of differentiation is unimportant).
The result is
true for $|\alpha| = 1$.
Then ${\partial \over \partial x_i}{\partial^\alpha n(x) \over \partial x^\alpha} = {\|x\|^2{\partial p_\alpha(x) \over \partial x_i} -(2k-1) x_i p_\alpha(x) \over \|x\|^{2 |\alpha|+1}} $, and so we see the result is true
for any index $\beta$ such that $|\beta|=|\alpha|+1$.
We note that since $\partial p_\alpha = |\alpha|$, we have
$| {\partial^\alpha n(x) \over \partial x^\alpha} | \le {K_\alpha \over \|x\|^{|\alpha|-1}}$ for some constant
$K_\alpha$.
The Faà di Bruno formula gives, for $x \neq 0$,
${\partial^\alpha \rho (x) \over \partial x^\alpha} = \sum_{\pi \in {\cal P}}
r^{(|\pi|)}(n(x))\prod_{\beta \in \pi} {\partial^\beta n (x) \over \partial x^\beta} $,
where ${\cal P}$ is the set of partitions of a $|\alpha|$ element subset of $1,...,n$ (and $|\alpha| \le k$, since $r$ is $C^k$).
(As an aside, to allow repetitions as in ${\partial^2 \over \partial x_i^2}$, strictly speaking we do not use partitions, but a modification
thereof that uses multi sets. The end result is the same, but involves even
more notation.)
We can use the above bound to get
\begin{eqnarray}
|{\partial^\alpha \rho (x) \over \partial x^\alpha}| &\le& \sum_{\pi \in {\cal P}}
| r^{(|\pi|)}(n(x)) | \prod_{\beta \in \pi} {K_\beta \over \|x\|^{|\beta|-1}} \\
&=& \sum_{\pi \in {\cal P}}
| r^{(|\pi|)}(n(x)) | { \prod_{\beta \in \pi} K_\beta \over \|x\|^{|\alpha|-|\pi|}} \\
&=&
\sum_{\pi \in {\cal P}} { | r^{(|\pi|)}(n(x)) | \over \|x\|^{|\alpha|-|\pi|}} \prod_{\beta \in \pi} K_\beta
\end{eqnarray}
Since $|\pi| \le |\alpha| \le k $, we see that
${ | r^{(|\pi|)}(n(x)) | \over \|x\|^{|\alpha|-|\pi|}} \to 0$ and so
${\partial^\alpha \rho (x) \over \partial x^\alpha} \to 0$.
Finally, we need to show that for $|\alpha| < k$, the function $x \mapsto {\partial^\alpha \rho (x) \over \partial x^\alpha}$ is differentiable with derivative zero.
In the following I am using the technical result for $r$ mentioned above:
Since $|\rho(x)-\rho(0) -0\cdot x| = |r(n(x))| \le \epsilon \|x\|$, we see that $\rho$ is differentiable with derivative zero, hence the result is true for $|\alpha|=1$. How suppose the result is true for $|\alpha| < k$, then,
using the fact that $|r^{(i)}(n(x))| \le \epsilon \|x\|^{k-i}$, we have
\begin{eqnarray}
|{\partial^\alpha \rho (x) \over \partial x^\alpha} - {\partial^\alpha \rho (0) \over \partial x^\alpha} - 0 \cdot x| &= & |{\partial^\alpha \rho (x) \over \partial x^\alpha}| \\
&\le& \sum_{\pi \in {\cal P}} { | r^{(|\pi|)}(n(x)) | \over \|x\|^{|\alpha|-|\pi|}} \prod_{\beta \in \pi} K_\beta \\
&\le& \epsilon \sum_{\pi \in {\cal P}} { \|x\|^{k-|\pi|} \over \|x\|^{|\alpha|-|\pi|}} \prod_{\beta \in \pi} K_\beta \\
= \epsilon \|x\|^{k-|\alpha|} \sum_{\pi \in {\cal P}} \prod_{\beta \in \pi} K_\beta
\end{eqnarray}
It follows from this that $x \mapsto {\partial^\alpha \rho (x) \over \partial x^\alpha}$ is differentiable at $x=0$ and has derivative zero, hence the result is true for derivatives of order $\beta$ such that $|\beta|=|\alpha|+1$.
